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Algorithm Design and Standard Methods: Question 9

Syllabus 7.4

Multiple choice 1 mark

An online store keeps the prices, in dollars, of 5 items in a one-dimensional (1D) array called Prices. The pseudocode algorithm below is run on this array.

DECLARE Prices : ARRAY[1:5] OF REAL
DECLARE Index : INTEGER
DECLARE Cheapest : REAL
Cheapest ← Prices[1]
FOR Index ← 2 TO 5
    IF Prices[Index] < Cheapest THEN
        Cheapest ← Prices[Index]
    ENDIF
NEXT Index
OUTPUT Cheapest

What does this algorithm calculate?

Choose an answer to check it, then compare with the worked solution below.
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Worked solution

Step 1: Identify what Cheapest starts as and how it changes

Cheapest is initialised to Prices[1], so it begins as one of the actual prices in the array, not zero and not a running total. Inside the loop, Cheapest is only ever replaced entirely by Prices[Index]. It is never added to and never divided.

Step 2: Trace the comparison being made

For each remaining item (Index from 2 to 5), the algorithm checks IF Prices[Index] < Cheapest THEN. Only when a price is smaller than the current value of Cheapest does the replacement Cheapest ← Prices[Index] happen. This means Cheapest can only ever get smaller (or stay the same) as the loop runs. It never increases.

Step 3: Rule out the other standard methods

  • Totalling (option A) would need Cheapest ← Cheapest + Prices[Index], adding every price together; here, values only replace Cheapest, they are never added to it.
  • Counting (option B) would need a separate counter variable incremented by 1 each time a condition is true; Cheapest always holds a full price value, not a count of comparisons.
  • Averaging (option D) would need a running total divided by 5 at the end; there is no division anywhere in this algorithm.

Step 4: Confirm what the algorithm actually finds

Because Cheapest starts at one price and is only ever overwritten by a smaller price found later in the array, after the loop finishes Cheapest holds the smallest of all 5 prices, the standard “finding a minimum” method, applied here to find the cheapest item.

Final answer

  • The algorithm finds the cheapest (lowest) price among the 5 items, option C.